Given data:
- It is a graphical display where the data is grouped in to ranges
- A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
- It is an accurate representation of the distribution of numerical data.
<em>From Figure:</em>
Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).
<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>
Answer:
the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm
Explanation:
Given data
Φ = 5.32 eV
to find out
the longest wavelength
solution
we know that
hf = k(maximum) +Ф ...............1
here we consider k(maximum ) will be zero because photon wavelength max when low photon energy
so hf = 0
and hc/ λ = +Ф
so λ = hc/Ф ................2
now put value hc = 1240 ev nm and Φ = 5.32 eV
so hc = 1240 / 5.32
hc = 233 nm
the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm
Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
Answer:
ФE = 9.403W
Explanation:
In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:
(1)
A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2
E: magnitude of the electric field = 95.0N/C
θ: angle between the direction of the electric field and the normal to the surface of the sheet
You replace the values of the parameters in the equation (1):
The magnitude of the electric flux is trough the sheet is 9.403W
